Periodic Trends — Practice · Grade 11 Science

Chapter: Atoms, Electrons & Periodicity

What this practice covers

By the end of Periodic Trends you should be able to:

10 questions, multiple choice, numeric and written answer, each with a worked explanation.

Vocabulary used: atomic radius, ionization energy, electronegativity, effective nuclear charge (Zeff), shielding (screening), periodic trend.

Problem 1

As you move left to right across a period, atomic radius generally…
  1. increases
  2. decreases
  3. stays the same
  4. doubles each time
Answer and explanation

decreases

Rising effective nuclear charge across a period pulls the electron cloud inward, so atoms get smaller.

Problem 2

Which of these elements has the highest electronegativity?
  1. chlorine (Cl)
  2. sodium (Na)
  3. fluorine (F)
  4. oxygen (O)
Answer and explanation

fluorine (F)

Fluorine, at the top right of the table (excluding noble gases), is the most electronegative element.

Problem 3

First ionization energy generally ___ as you move down a group.
  1. stays the same
  2. becomes negative
  3. increases
  4. decreases
Answer and explanation

decreases

Down a group, the outer electron is farther out and better shielded, so it is easier to remove — ionization energy decreases.

Problem 4

A neutral fluorine atom has 9 protons and 2 core (inner-shell) electrons. Using the simple approximation Zeff ≈ (protons) − (core electrons), estimate the effective nuclear charge felt by a valence electron.
Answer and explanation

7

Zeff ≈ 9 − 2 = 7. The 7 valence electrons feel a strong pull, which is why fluorine is so small and so electronegative.

Problem 5

Explain, in terms of effective nuclear charge, why atomic radius decreases across a period even though electrons are being added.
Answer and explanation

Across a period, protons are added to the nucleus while the new electrons enter the same energy level, adding little shielding. Effective nuclear charge rises, so the stronger pull draws the whole electron cloud inward and the atom shrinks.

Problem 6

Atomic radius and ionization energy show opposite trends. An atom with a very small radius will therefore tend to have a…
  1. high ionization energy
  2. low ionization energy
  3. zero ionization energy
  4. negative ionization energy
Answer and explanation

high ionization energy

A small radius means the outer electron is close to the nucleus and held tightly, so it takes a lot of energy to remove — a high ionization energy.

Problem 7

Compare Na and Na⁺ size qualitatively. Show the reasoning, not just the conclusion.
Hints

Trend statements require a specified property and comparable definition.

Answer and explanation

Na⁺ is smaller; losing the outer electron removes the occupied outer level and changes electron interactions.

Problem 8

Why are periodic trends useful but not a guarantee of perfect monotonic behavior?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Orbital structure, electron pairing and measurement definitions produce exceptions; consult data for precise comparisons.

Problem 9

Equal electron count isolates an important nuclear-charge comparisonEqual electron count isolates an important nuclear-charge comparisonElectron countO²⁻: 8 protons10F⁻: 9 protons10Na⁺: 11 protons10Same electron count; increasing nuclear chargeUsual isoelectronic radius order: O²⁻ > F⁻ > Na⁺
Original equal-electron inventory. Bar lengths show electron counts only, not ionic radii.
Interpret the supplied visual. Which is smallest in the displayed isoelectronic series, and which variable supports the comparison?
Answer and explanation

Na⁺ is smallest in the usual comparison; all three have ten electrons but Na⁺ has the largest proton count, 11.

Isoelectronic species have the same electron count under the compared state. O²⁻, F⁻ and Na⁺ each have ten electrons, but their proton counts rise from eight to eleven. Within this selected series, stronger nuclear attraction generally contracts the electron distribution, giving O²⁻ the larger and Na⁺ the smaller radius under compatible definitions. This is different from saying every positive ion of every element is smaller than every negative ion. Match the electron count and comparison conditions first.

Na⁺ is smallest in the usual comparison; all three have ten electrons but Na⁺ has the largest proton count, 11.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 10

Rank Ne, F⁻ and Mg²⁺ by the usual radius trend for an isoelectronic series. Give electron/proton counts and the model limitation.
Answer and explanation

All have ten electrons; proton counts are F=9, Ne=10, Mg=12. The usual order is F⁻ > Ne > Mg²⁺ under a compatible radius comparison. Defined atomic/ionic radii and environments must be matched before treating reference numbers as directly interchangeable.

All have ten electrons; proton counts are F=9, Ne=10, Mg=12. The usual order is F⁻ > Ne > Mg²⁺ under a compatible radius comparison. Defined atomic/ionic radii and environments must be matched before treating reference numbers as directly interchangeable.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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